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Match the following: Column I Column II A. \(\mathrm{H}_{2} \mathrm{SO}_{5}\) (p) \(+6\) oxidation state B. \(\mathrm{CrO}_{5}\) (q) peroxy linkage C. \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) (r) Oxidant D. \(\mathrm{H}_{2} \mathrm{O}_{2}\) (s) Bleaching action

Short Answer

Expert verified
A matches with q and p, B with q and r, C with r, D with s.

Step by step solution

01

Analyze  his{H}_{2}{SO}_{5}

Peroxomonosulfuric acid ( his{H}_{2}{SO}_{5}) contains a peroxy linkage (p=O) as well as sulfur in its +6 oxidation state. Therefore, match it with (q) peroxy linkage and (p).
02

Examine  his{CrO}_{5}

Chromium pentaoxide ( his{CrO}_{5}) has chromium in the +6 oxidation state and contains peroxide linkages. Its striking color changes and reactivity are due to its oxidizing nature, which matches it with (q) peroxy linkage and (r) an oxidant.
03

Evaluate  his{K}_{2}{Cr}_{2}{O}_{7}

Potassium dichromate ( his{K}_{2}{Cr}_{2}{O}_{7}) is known for being a strong oxidizing agent in acidic conditions. Therefore, match it with (r), oxidant.
04

Consider  his{H}_{2}{O}_{2}

Hydrogen peroxide ( his{H}_{2}{O}_{2}) is commonly used as a bleaching agent due its release of oxygen free radicals that help in oxidation. Therefore, match it with (s), bleaching action.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Peroxy Linkage
Peroxy linkage is an interesting chemical concept that often shows up in compounds such as peroxides. It is essentially characterized by an oxygen-oxygen single bond, represented as \[ \text{-O-O-} \]This linkage is quite reactive due to the relatively weak bond between the two oxygen atoms. In particular, - Peroxomonosulfuric acid (\(\text{H}_2\text{SO}_5\)) contains a peroxy bond which contributes to its ability to participate in oxidation reactions.
- Chromium pentaoxide (\(\text{CrO}_5\)) also contains peroxy linkages, which plays a crucial role in enhancing its reactivity. The presence of peroxy linkage in a compound significantly affects its chemical behavior, making it more reactive by facilitating the release of oxygen species, which can further act as oxidizing agents.
Oxidant
An oxidant, also known as an oxidizing agent, is a substance that has the ability to oxidize other substances. It means that the oxidant itself gets reduced while the other substance undergoes oxidation. This process is essential in many chemical reactions, particularly in redox (reduction-oxidation) reactions. - For example, Potassium dichromate (\(\text{K}_2\text{Cr}_2\text{O}_7\)) is a well-known strong oxidizing agent, especially effective in acidic solutions, where it participates in various industrial and laboratory processes.
- Chromium pentaoxide (\(\text{CrO}_5\)) is another oxidant due to its ability to change oxidation states and participate in oxidative processes. In essence, the oxidant plays a critical role in driving chemical reactions across multiple industries by transferring oxygen atoms or accepting electrons from other species.
Bleaching Action
Bleaching action is a process commonly applied to remove color from materials. It involves oxidative reactions that break down colored substances, rendering them colorless. Hydrogen peroxide (\(\text{H}_2\text{O}_2\)) is a classic example of a bleaching agent. Its effectiveness stems from its ability to release oxygen radicals.Here's how it works: - During bleaching, \(\text{H}_2\text{O}_2\) decomposes to produce oxygen radicals, highly reactive species that can disrupt the structure of colored compounds.
- These radicals target the chromophores, the parts of molecules responsible for color, breaking them down into smaller, non-colored components. The bleaching process is widely used in both industrial applications such as textile, paper, and pulp industries, as well as in household products for clothes and teeth whitening. It's an excellent example of an application where chemical principles power practical uses.

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Most popular questions from this chapter

Resistance of a conductivity cell filled with a solution of an electrolyte of concentration \(0.1 \mathrm{M}\) is 100 \(\Omega\). The conductivity of this solution is \(1.29 \mathrm{~S} \mathrm{~m}^{-1}\). Resistance of the same cell when filled with \(0.2 \mathrm{M}\) of the same solution is \(520 \Omega\). The molar conductivity of \(0.02 \mathrm{M}\) solution of the electrolyte will be a. \(124 \times 10^{-4} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1}\) b. \(1240 \times 10^{-4} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1}\) c. \(1.24 \times 10^{-4} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1}\) d. \(12.4 \times 10^{-4} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1}\)

The \(\mathrm{emf}, \mathrm{E}\), is related to the change in Gibbs free energy, \(\Delta \mathrm{G}: \Delta \mathrm{G}=-\mathrm{nFE}\), where is the number of electrons transferred during the redox process and \(F\) is a unit called the Faraday. The faraday is the amount of charge on \(1 \mathrm{~mol}\) of electrons: \(1 \mathrm{~F}=96,500 \mathrm{C} / \mathrm{mol}\). Because \(\mathrm{E}\) is related to \(\Delta \mathrm{G}\), the sign of \(\mathrm{E}\) indicates whether a redox process is spontaneous: \(\mathrm{E}>0\) indicates a spontaneous process, and \(\mathrm{E}<0\) indicates a non-spontaneous one. Given the cell: \(\mathrm{Cd}(\mathrm{s})\left|\mathrm{Cd}(\mathrm{OH})_{2}(\mathrm{~s})\right| \mathrm{NaOH}(\mathrm{aq}, 0.01 \mathrm{M})\) \(\mid \mathrm{H}_{2}(\mathrm{~g}, 1\) bar \(\mid \mathrm{Pt}(\mathrm{s}))\)with \(\mathrm{E}_{\text {cell }}=0.0 \mathrm{~V}\). If \(\mathrm{E}_{\mathrm{Cd}}^{\circ}{ }_{\mathrm{Cd}}^{2+}=-0.39 \mathrm{~V}\), then \(\mathrm{K}_{\mathrm{sp}}\) of \(\mathrm{Cd}\left(\mathrm{OH}_{2}\right)\) is: a. \(10^{-15}\) b. \(10^{-13}\) c. \(10^{13}\) d. \(10^{15}\)

Excess of KI reacts with \(\mathrm{CuSO}_{4}\) solution and then \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) solution is added to it. Which of the statements is/are correct for this reaction? a. \(\mathrm{Cu}_{2} \mathrm{I}_{2}\) is formed b. \(\mathrm{CuI}_{2}\) is formed c. \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) is oxidized d. evolved \(\mathrm{I}_{2}\) is reduced

The correct order of equivalent conductance at infinite dilution of \(\mathrm{LiCl}, \mathrm{NaCl}\) and \(\mathrm{KCl}\) is: a. \(\mathrm{LiCl}>\mathrm{NaCl}>\mathrm{KCl}\) b. \(\mathrm{KCl}>\mathrm{NaCl}>\mathrm{LiCl}\) c. \(\mathrm{NaCl}>\mathrm{KCl}>\mathrm{LiCl}\) d. \(\mathrm{LiCl}>\mathrm{KCl}>\mathrm{NaCl}\)

On the basis of the following \(\mathrm{E}^{\circ}\) values, the strongest oxidizing agent is: \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-} \rightarrow\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}, \mathrm{E}^{\circ}=-0.35 \mathrm{~V}\) \(\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}+\mathrm{e}^{-}, \mathrm{E}^{\circ}=-0.77 \mathrm{~V}\) a. \(\mathrm{Fe}^{3+}\) b. \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) c. \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\) d. \(\mathrm{Fe}^{2+}\)

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